HDOJ: 1009FatMouse' Trade

Problem Description


FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean. The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does
not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can
obtain.

Input：

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's.
All integers are not greater than 1000.

Output
：

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

Sample Input:

5 3 （5个，有3个交换的地方）

7 2 （2换7）

4 3

5 2

20 3

25 18

24 15

15 10

-1 -1

Sample Output;

13.333

31.500

解答：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
double c;
double d;
}s[10001];
bool cmp(node ai,node bi)
{
return ai.c/ai.d>bi.c/bi.d;
}
int main ()
{
double  a;
int b;
int i;
double p;
while(scanf("%lf%d",&a,&b)!=EOF)
{p=0.0;
if(a==-1&&b==-1)
break;
for(i=0;i<b;i++)
scanf("%lf%lf",&s[i].c,&s[i].d);
sort(s,s+b,cmp);
for(i=0;a-s[i].d>=0;i++)
{
a=a-s[i].d;
p=p+s[i].c;
}
p=p+s[i].c*a/s[i].d;
printf("%.3f\n",p);
}
return 0;
}